Internal problem ID [6497]
Internal file name [OUTPUT/5745_Sunday_June_05_2022_03_52_29_PM_25435504/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Problems for review and
discovert. (B) Challenge Problems . Page 194
Problem number: 1(b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {9 \left (x -2\right )^{2} \left (x -3\right ) y^{\prime \prime }+6 x \left (x -2\right ) y^{\prime }+16 y=0} \] With the expansion point for the power series method at \(x = \infty \).
Since expansion is around \(\infty \), then the independent variable \(x\) is replaced by \(\frac {1}{t}\) and the expansion is made around \(t = 0\) and after solving, the solution is changed back to \(x\) using \(x = \frac {1}{t}\). Changing variables results in the new ode \begin {align*} -9 \left (-2+\frac {1}{t}\right )^{2} \left (\frac {1}{t}-3\right ) \left (-\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-2 t \left (\frac {d}{d t}y \left (t \right )\right )\right ) t^{2}-\left (\frac {6}{t^{2}}-\frac {12}{t}\right ) \left (\frac {d}{d t}y \left (t \right )\right ) t^{2}+16 y \left (t \right ) = 0 \end {align*}
The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (-108 t^{4}+144 t^{3}-63 t^{2}+9 t \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (-216 t^{3}+288 t^{2}-114 t +12\right ) \left (\frac {d}{d t}y \left (t \right )\right )+16 y \left (t \right ) = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d t^{2}}y \left (t \right )+p(t) \frac {d}{d t}y \left (t \right ) + q(t) y \left (t \right ) &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {12 t^{2}-10 t +\frac {4}{3}}{t \left (-1+3 t \right ) \left (2 t -1\right )}\\ q(t) &= -\frac {16}{9 \left (2 t -1\right )^{2} \left (-1+3 t \right ) t}\\ \end {align*}
| \(p(t)=\frac {12 t^{2}-10 t +\frac {4}{3}}{t \left (-1+3 t \right ) \left (2 t -1\right )}\) | |
| singularity | type |
| \(t = 0\) | \(\text {``regular''}\) |
| \(t = {\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(t = {\frac {1}{3}}\) | \(\text {``regular''}\) |
| \(q(t)=-\frac {16}{9 \left (2 t -1\right )^{2} \left (-1+3 t \right ) t}\) | |
| singularity | type |
| \(t = 0\) | \(\text {``regular''}\) |
| \(t = {\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(t = {\frac {1}{3}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, {\frac {1}{2}}, {\frac {1}{3}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -9 t \left (12 t^{3}-16 t^{2}+7 t -1\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (-216 t^{3}+288 t^{2}-114 t +12\right ) \left (\frac {d}{d t}y \left (t \right )\right )+16 y \left (t \right ) = 0 \] Let the solution be represented as Frobenius power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} -9 t \left (12 t^{3}-16 t^{2}+7 t -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (-216 t^{3}+288 t^{2}-114 t +12\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+16 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-108 t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}144 t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-63 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-216 t^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}288 t^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-114 t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 \left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-108 t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-108 a_{n -3} \left (n +r -3\right ) \left (n -4+r \right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}144 t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}144 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-63 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-63 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-216 t^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-216 a_{n -3} \left (n +r -3\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}288 t^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}288 a_{n -2} \left (n +r -2\right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-114 t^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-114 a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}16 a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} t^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =3}{\sum }}\left (-108 a_{n -3} \left (n +r -3\right ) \left (n -4+r \right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}144 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-63 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-216 a_{n -3} \left (n +r -3\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}288 a_{n -2} \left (n +r -2\right ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-114 a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 \left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+12 \left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 9 t^{-1+r} a_{0} r \left (-1+r \right )+12 r a_{0} t^{-1+r} = 0 \] Or \[ \left (9 t^{-1+r} r \left (-1+r \right )+12 r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}+3 r \right ) t^{-1+r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 9 r^{2}+3 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}+3 r \right ) t^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{3}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -\frac {1}{3}} \end {align*}
We start by finding \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} -108 a_{n -3} \left (n +r -3\right ) \left (n -4+r \right )+144 a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )-63 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 a_{n} \left (n +r \right ) \left (n +r -1\right )-216 a_{n -3} \left (n +r -3\right )+288 a_{n -2} \left (n +r -2\right )-114 a_{n -1} \left (n +r -1\right )+12 a_{n} \left (n +r \right )+16 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {108 n^{2} a_{n -3}-144 n^{2} a_{n -2}+63 n^{2} a_{n -1}+216 n r a_{n -3}-288 n r a_{n -2}+126 n r a_{n -1}+108 r^{2} a_{n -3}-144 r^{2} a_{n -2}+63 r^{2} a_{n -1}-540 n a_{n -3}+432 n a_{n -2}-75 n a_{n -1}-540 r a_{n -3}+432 r a_{n -2}-75 r a_{n -1}+648 a_{n -3}-288 a_{n -2}-4 a_{n -1}}{9 n^{2}+18 n r +9 r^{2}+3 n +3 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (108 a_{n -3}-144 a_{n -2}+63 a_{n -1}\right ) n^{2}+\left (-540 a_{n -3}+432 a_{n -2}-75 a_{n -1}\right ) n +648 a_{n -3}-288 a_{n -2}-4 a_{n -1}}{9 n^{2}+3 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {3004}{405}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
| \(a_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {3004}{405}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=-{\frac {285704}{15795}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
| \(a_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {3004}{405}}\) |
| \(a_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {285704}{15795}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {822592}{18225}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
| \(a_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {3004}{405}}\) |
| \(a_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {285704}{15795}}\) |
| \(a_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {822592}{18225}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )} \] Which for the root \(r = 0\) becomes \[ a_{6}=-{\frac {4666732192}{40514175}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
| \(a_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {3004}{405}}\) |
| \(a_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {285704}{15795}}\) |
| \(a_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {822592}{18225}}\) |
| \(a_{6}\) | \(\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )}\) | \(-{\frac {4666732192}{40514175}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {80674338123 r^{14}+4080569276151 r^{13}+92307128293827 r^{12}+1233512636444739 r^{11}+10829721106831365 r^{10}+65685074939730909 r^{9}+281466610429694193 r^{8}+855089251792429041 r^{7}+1814849020228964364 r^{6}+2584390104306680040 r^{5}+2241655427586657456 r^{4}+848621079979968528 r^{3}-251613495711864000 r^{2}-338590375929007104 r -80579334056574976}{2187 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right ) \left (3 r^{2}+43 r +154\right )} \] Which for the root \(r = 0\) becomes \[ a_{7}=-{\frac {401483448544}{1336967775}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {28}{9}}\) |
| \(a_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {3004}{405}}\) |
| \(a_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {285704}{15795}}\) |
| \(a_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {822592}{18225}}\) |
| \(a_{6}\) | \(\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )}\) | \(-{\frac {4666732192}{40514175}}\) |
| \(a_{7}\) | \(\frac {80674338123 r^{14}+4080569276151 r^{13}+92307128293827 r^{12}+1233512636444739 r^{11}+10829721106831365 r^{10}+65685074939730909 r^{9}+281466610429694193 r^{8}+855089251792429041 r^{7}+1814849020228964364 r^{6}+2584390104306680040 r^{5}+2241655427586657456 r^{4}+848621079979968528 r^{3}-251613495711864000 r^{2}-338590375929007104 r -80579334056574976}{2187 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right ) \left (3 r^{2}+43 r +154\right )}\) | \(-{\frac {401483448544}{1336967775}}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}+a_{7} t^{7}+a_{8} t^{8}\dots \\ &= 1-\frac {4 t}{3}-\frac {28 t^{2}}{9}-\frac {3004 t^{3}}{405}-\frac {285704 t^{4}}{15795}-\frac {822592 t^{5}}{18225}-\frac {4666732192 t^{6}}{40514175}-\frac {401483448544 t^{7}}{1336967775}+O\left (t^{8}\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} -108 b_{n -3} \left (n +r -3\right ) \left (n -4+r \right )+144 b_{n -2} \left (n +r -2\right ) \left (n +r -3\right )-63 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 b_{n} \left (n +r \right ) \left (n +r -1\right )-216 b_{n -3} \left (n +r -3\right )+288 b_{n -2} \left (n +r -2\right )-114 b_{n -1} \left (n +r -1\right )+12 \left (n +r \right ) b_{n}+16 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {108 n^{2} b_{n -3}-144 n^{2} b_{n -2}+63 n^{2} b_{n -1}+216 n r b_{n -3}-288 n r b_{n -2}+126 n r b_{n -1}+108 r^{2} b_{n -3}-144 r^{2} b_{n -2}+63 r^{2} b_{n -1}-540 n b_{n -3}+432 n b_{n -2}-75 n b_{n -1}-540 r b_{n -3}+432 r b_{n -2}-75 r b_{n -1}+648 b_{n -3}-288 b_{n -2}-4 b_{n -1}}{9 n^{2}+18 n r +9 r^{2}+3 n +3 r}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{n} = \frac {\left (108 b_{n -3}-144 b_{n -2}+63 b_{n -1}\right ) n^{2}+\left (-612 b_{n -3}+528 b_{n -2}-117 b_{n -1}\right ) n +840 b_{n -3}-448 b_{n -2}+28 b_{n -1}}{9 n^{2}-3 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{3}=-{\frac {7781}{810}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
| \(b_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {7781}{810}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{4}=-{\frac {22151}{1215}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
| \(b_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {7781}{810}}\) |
| \(b_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {22151}{1215}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{5}=-{\frac {669229}{18225}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
| \(b_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {7781}{810}}\) |
| \(b_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {22151}{1215}}\) |
| \(b_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {669229}{18225}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{6}=-{\frac {216463313}{2788425}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
| \(b_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {7781}{810}}\) |
| \(b_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {22151}{1215}}\) |
| \(b_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {669229}{18225}}\) |
| \(b_{6}\) | \(\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )}\) | \(-{\frac {216463313}{2788425}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {80674338123 r^{14}+4080569276151 r^{13}+92307128293827 r^{12}+1233512636444739 r^{11}+10829721106831365 r^{10}+65685074939730909 r^{9}+281466610429694193 r^{8}+855089251792429041 r^{7}+1814849020228964364 r^{6}+2584390104306680040 r^{5}+2241655427586657456 r^{4}+848621079979968528 r^{3}-251613495711864000 r^{2}-338590375929007104 r -80579334056574976}{2187 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right ) \left (3 r^{2}+43 r +154\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{7}=-{\frac {7179886604}{41826375}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {63 r^{2}+51 r -16}{9 r^{2}+21 r +12}\) | \(-{\frac {13}{3}}\) |
| \(b_{2}\) | \(\frac {2673 r^{4}+10044 r^{3}+9441 r^{2}+438 r -1568}{81 r^{4}+540 r^{3}+1305 r^{2}+1350 r +504}\) | \(-{\frac {251}{45}}\) |
| \(b_{3}\) | \(\frac {95499 r^{6}+844911 r^{5}+2671137 r^{4}+3571317 r^{3}+1572804 r^{2}-419508 r -336448}{729 r^{6}+9477 r^{5}+49815 r^{4}+135135 r^{3}+198936 r^{2}+150228 r +45360}\) | \(-{\frac {7781}{810}}\) |
| \(b_{4}\) | \(\frac {3103353 r^{8}+49898592 r^{7}+320994738 r^{6}+1054759968 r^{5}+1851078825 r^{4}+1583317152 r^{3}+348879276 r^{2}-293260704 r -127995392}{81 \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {22151}{1215}}\) |
| \(b_{5}\) | \(\frac {95127939 r^{10}+2420949951 r^{9}+26064871578 r^{8}+154759665150 r^{7}+552318920667 r^{6}+1204450764687 r^{5}+1536667458552 r^{4}+964786860468 r^{3}+52458046992 r^{2}-235978209024 r -76652412928}{243 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right )}\) | \(-{\frac {669229}{18225}}\) |
| \(b_{6}\) | \(\frac {2806539921 r^{12}+103709648268 r^{11}+1676277992835 r^{10}+15583794209910 r^{9}+92090095190235 r^{8}+360297425137320 r^{7}+940035462956685 r^{6}+1596754328979150 r^{5}+1637397360960804 r^{4}+792733805144712 r^{3}-86042236842144 r^{2}-247522178677248 r -66902272704512}{729 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right )}\) | \(-{\frac {216463313}{2788425}}\) |
| \(b_{7}\) | \(\frac {80674338123 r^{14}+4080569276151 r^{13}+92307128293827 r^{12}+1233512636444739 r^{11}+10829721106831365 r^{10}+65685074939730909 r^{9}+281466610429694193 r^{8}+855089251792429041 r^{7}+1814849020228964364 r^{6}+2584390104306680040 r^{5}+2241655427586657456 r^{4}+848621079979968528 r^{3}-251613495711864000 r^{2}-338590375929007104 r -80579334056574976}{2187 \left (3 r^{2}+31 r +80\right ) \left (3 r^{2}+25 r +52\right ) \left (27 r^{6}+351 r^{5}+1845 r^{4}+5005 r^{3}+7368 r^{2}+5564 r +1680\right ) \left (3 r^{2}+37 r +114\right ) \left (3 r^{2}+43 r +154\right )}\) | \(-{\frac {7179886604}{41826375}}\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= 1 \left (b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}+b_{7} t^{7}+b_{8} t^{8}\dots \right ) \\ &= \frac {1-\frac {13 t}{3}-\frac {251 t^{2}}{45}-\frac {7781 t^{3}}{810}-\frac {22151 t^{4}}{1215}-\frac {669229 t^{5}}{18225}-\frac {216463313 t^{6}}{2788425}-\frac {7179886604 t^{7}}{41826375}+O\left (t^{8}\right )}{t^{\frac {1}{3}}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (1-\frac {4 t}{3}-\frac {28 t^{2}}{9}-\frac {3004 t^{3}}{405}-\frac {285704 t^{4}}{15795}-\frac {822592 t^{5}}{18225}-\frac {4666732192 t^{6}}{40514175}-\frac {401483448544 t^{7}}{1336967775}+O\left (t^{8}\right )\right ) + \frac {c_{2} \left (1-\frac {13 t}{3}-\frac {251 t^{2}}{45}-\frac {7781 t^{3}}{810}-\frac {22151 t^{4}}{1215}-\frac {669229 t^{5}}{18225}-\frac {216463313 t^{6}}{2788425}-\frac {7179886604 t^{7}}{41826375}+O\left (t^{8}\right )\right )}{t^{\frac {1}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} \left (1-\frac {4 t}{3}-\frac {28 t^{2}}{9}-\frac {3004 t^{3}}{405}-\frac {285704 t^{4}}{15795}-\frac {822592 t^{5}}{18225}-\frac {4666732192 t^{6}}{40514175}-\frac {401483448544 t^{7}}{1336967775}+O\left (t^{8}\right )\right )+\frac {c_{2} \left (1-\frac {13 t}{3}-\frac {251 t^{2}}{45}-\frac {7781 t^{3}}{810}-\frac {22151 t^{4}}{1215}-\frac {669229 t^{5}}{18225}-\frac {216463313 t^{6}}{2788425}-\frac {7179886604 t^{7}}{41826375}+O\left (t^{8}\right )\right )}{t^{\frac {1}{3}}} \\ \end{align*} Replacing \(t\) by \(\frac {1}{x}\) gives \begin {align*} y = c_{1} \left (1-\frac {4}{3 x}-\frac {28}{9 x^{2}}-\frac {3004}{405 x^{3}}-\frac {285704}{15795 x^{4}}-\frac {822592}{18225 x^{5}}-\frac {4666732192}{40514175 x^{6}}-\frac {401483448544}{1336967775 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+\frac {c_{2} \left (1-\frac {13}{3 x}-\frac {251}{45 x^{2}}-\frac {7781}{810 x^{3}}-\frac {22151}{1215 x^{4}}-\frac {669229}{18225 x^{5}}-\frac {216463313}{2788425 x^{6}}-\frac {7179886604}{41826375 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )}{\left (\frac {1}{x}\right )^{\frac {1}{3}}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {4}{3 x}-\frac {28}{9 x^{2}}-\frac {3004}{405 x^{3}}-\frac {285704}{15795 x^{4}}-\frac {822592}{18225 x^{5}}-\frac {4666732192}{40514175 x^{6}}-\frac {401483448544}{1336967775 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+\frac {c_{2} \left (1-\frac {13}{3 x}-\frac {251}{45 x^{2}}-\frac {7781}{810 x^{3}}-\frac {22151}{1215 x^{4}}-\frac {669229}{18225 x^{5}}-\frac {216463313}{2788425 x^{6}}-\frac {7179886604}{41826375 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )}{\left (\frac {1}{x}\right )^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {4}{3 x}-\frac {28}{9 x^{2}}-\frac {3004}{405 x^{3}}-\frac {285704}{15795 x^{4}}-\frac {822592}{18225 x^{5}}-\frac {4666732192}{40514175 x^{6}}-\frac {401483448544}{1336967775 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+\frac {c_{2} \left (1-\frac {13}{3 x}-\frac {251}{45 x^{2}}-\frac {7781}{810 x^{3}}-\frac {22151}{1215 x^{4}}-\frac {669229}{18225 x^{5}}-\frac {216463313}{2788425 x^{6}}-\frac {7179886604}{41826375 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )}{\left (\frac {1}{x}\right )^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 \left (x -2\right )^{2} \left (x -3\right ) y^{\prime \prime }+\left (6 x^{2}-12 x \right ) y^{\prime }+16 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {16 y}{9 \left (x -2\right )^{2} \left (x -3\right )}-\frac {2 y^{\prime } x}{3 \left (x -3\right ) \left (x -2\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {2 y^{\prime } x}{3 \left (x -3\right ) \left (x -2\right )}+\frac {16 y}{9 \left (x -2\right )^{2} \left (x -3\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {2 x}{3 \left (x -3\right ) \left (x -2\right )}, P_{3}\left (x \right )=\frac {16}{9 \left (x -2\right )^{2} \left (x -3\right )}\right ] \\ {} & \circ & \left (x -2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =2 \\ {} & {} & \left (\left (x -2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}2}}}=-\frac {4}{3} \\ {} & \circ & \left (x -2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =2 \\ {} & {} & \left (\left (x -2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}2}}}=-\frac {16}{9} \\ {} & \circ & x =2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 \left (x -2\right )^{2} \left (x -3\right ) y^{\prime \prime }+6 x \left (x -2\right ) y^{\prime }+16 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (9 u^{3}-9 u^{2}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (6 u^{2}+12 u \right ) \left (\frac {d}{d u}y \left (u \right )\right )+16 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (9 r^{2}-21 r -16\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (9 k^{2}+18 k r +9 r^{2}-21 k -21 r -16\right )+3 a_{k -1} \left (k +r -1\right ) \left (3 k -4+3 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -9 r^{2}+21 r +16=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {7}{6}-\frac {\sqrt {113}}{6}, \frac {7}{6}+\frac {\sqrt {113}}{6}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (k +r -1\right ) \left (k -\frac {4}{3}+r \right ) a_{k -1}-9 \left (k^{2}+\left (2 r -\frac {7}{3}\right ) k +r^{2}-\frac {7 r}{3}-\frac {16}{9}\right ) a_{k}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 9 \left (k +r \right ) \left (k -\frac {1}{3}+r \right ) a_{k}-9 \left (\left (k +1\right )^{2}+\left (2 r -\frac {7}{3}\right ) \left (k +1\right )+r^{2}-\frac {7 r}{3}-\frac {16}{9}\right ) a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {3 \left (k +r \right ) \left (3 k +3 r -1\right ) a_{k}}{9 k^{2}+18 k r +9 r^{2}-3 k -3 r -28} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{6}-\frac {\sqrt {113}}{6} \\ {} & {} & a_{k +1}=\frac {3 \left (k +\frac {7}{6}-\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}-\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}+\frac {\sqrt {113}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{6}-\frac {\sqrt {113}}{6} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {7}{6}-\frac {\sqrt {113}}{6}}, a_{k +1}=\frac {3 \left (k +\frac {7}{6}-\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}-\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}+\frac {\sqrt {113}}{2}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -2\right )^{k +\frac {7}{6}-\frac {\sqrt {113}}{6}}, a_{k +1}=\frac {3 \left (k +\frac {7}{6}-\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}-\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}+\frac {\sqrt {113}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{6}+\frac {\sqrt {113}}{6} \\ {} & {} & a_{k +1}=\frac {3 \left (k +\frac {7}{6}+\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}+\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}-\frac {\sqrt {113}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{6}+\frac {\sqrt {113}}{6} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {7}{6}+\frac {\sqrt {113}}{6}}, a_{k +1}=\frac {3 \left (k +\frac {7}{6}+\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}+\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}-\frac {\sqrt {113}}{2}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -2\right )^{k +\frac {7}{6}+\frac {\sqrt {113}}{6}}, a_{k +1}=\frac {3 \left (k +\frac {7}{6}+\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}+\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}-\frac {\sqrt {113}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -2\right )^{k +\frac {7}{6}-\frac {\sqrt {113}}{6}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x -2\right )^{k +\frac {7}{6}+\frac {\sqrt {113}}{6}}\right ), a_{k +1}=\frac {3 \left (k +\frac {7}{6}-\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}-\frac {\sqrt {113}}{2}\right ) a_{k}}{9 k^{2}+18 k \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}-\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}+\frac {\sqrt {113}}{2}}, b_{k +1}=\frac {3 \left (k +\frac {7}{6}+\frac {\sqrt {113}}{6}\right ) \left (3 k +\frac {5}{2}+\frac {\sqrt {113}}{2}\right ) b_{k}}{9 k^{2}+18 k \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )+9 \left (\frac {7}{6}+\frac {\sqrt {113}}{6}\right )^{2}-3 k -\frac {63}{2}-\frac {\sqrt {113}}{2}}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 414
Order:=8; dsolve(9*(x-2)^2*(x-3)*diff(y(x),x$2)+6*x*(x-2)*diff(y(x),x)+16*y(x)=0,y(x),type='series',x=infinity);
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 130
AsymptoticDSolveValue[9*(x-2)^2*(x-3)*y''[x]+6*x*(x-2)*y'[x]+16*y[x]==0,y[x],{x,infinity,7}]
\[ y(x)\to c_2 \left (-\frac {13}{3 x^{2/3}}-\frac {251}{45 x^{5/3}}-\frac {7781}{810 x^{8/3}}-\frac {22151}{1215 x^{11/3}}-\frac {669229}{18225 x^{14/3}}-\frac {216463313}{2788425 x^{17/3}}-\frac {7179886604}{41826375 x^{20/3}}+\sqrt [3]{x}\right )+c_1 \left (-\frac {401483448544}{1336967775 x^7}-\frac {4666732192}{40514175 x^6}-\frac {822592}{18225 x^5}-\frac {285704}{15795 x^4}-\frac {3004}{405 x^3}-\frac {28}{9 x^2}-\frac {4}{3 x}+1\right ) \]